CE322 Basic Hydrology
Jorge A. Ramírez
Midterm Exam No. 2 - Solution
Problem 1 (50 points) During the month of August over a large tropical area, the net available energy flux Qn is 200 W m-2; the mean air temperature Ta is 21 ºC; the mean surface temperature To is 25 ºC; the relative humidity is 45%; and the mean wind speed is 2 m s-1 at z=2 m. At the above temperatures D/g is equal to 2.944. Assume zo=0.0001 m and p=100000 Pa. State any other additional assumptions.
a) Obtain the evaporation rate in millimeters per day using the Thornthwaite-Holzmann aerodynamic or mass transfer method, EA.
Obtain density of ambient air:
ra = (105 Pa)/(287
J/kg/K)/(294.15 K) = 1.1845 kg/m3
Obtain saturation vapor pressure of the
air using Clausius-Clapeyron equation:
es(Ta) = 611exp((-2.5 106/461.5)(1/294.15-1/273.15))
= 2521.18 Pa
Obtain actual vapor pressure of ambient
air:
ea
= 0.45 es(Ta)
= (0.5) 2364 Pa = 1134.53 Pa
EA = (0.622) (1.1845 kg/m3) (0.4)2 (2 m/s) 2521.18 Pa
(1-0.45)/105 Pa/(ln(2 m/0.0001
m))2
EA = 3.33 10-5 kg/m2/s
Expressed in mm/day:
EA = 3.33 10-5 (kg/m2/s) (10-3 m3/kg) (103 mm/m) (86400 s/day) = 2.88 mm/day
b) Obtain the equilibrium evaporation rate, Ee.
D/(D+g)
= (D/g)/(1+D/g) = 2.94/3.94 = 0.746
E
= 0.746 (200 W/m2)/(2.5 106
J/kg) = 5.97 10-5 kg/m2/s
Expressed in mm/day:
EA = 5.97 10-5 (kg/m2/s) (10-3 m3/kg) (103 mm/m) (86400 s/day) = 5.16 mm/day
c) Obtain an estimate of the evaporation rate over this region using Penman equation.
Penman
Equation:
g/(D+g)
= (g/g)/(1+D/g) = 1/3.94 = 0.254
E
= 0.746 (200 W/m2)/(2.5 106
J/kg) + 0.254 (3.33 10-5 kg/m2/s) = 6.813 10-5
kg/m2/s
d) Neglect the energy advected by the evaporated water, and obtain an estimate of the evaporation rate using the energy balance method with Bowen Ratio.
Neglecting
energy advected by evaporated water:
Obtain the Bowen Ratio, B:
Obtain saturation vapor pressure at the
surface temperature using Clausius-Clapeyron equation:
es(To) = 611exp((-2.5 106/461.5)(1/298.15-1/273.15))
= 3222.85 Pa
B
= g (298.15 – 294.15) K/(3222.85
– 2521.18) Pa = 0.368
Finally,
E = (200 W/m2)/(2.5 106
J/kg)/(1 + 0.368) = 5.848 10-5
kg/m2/s
Problem 2 (25 points) Using the Strahler ordering scheme the order W of a watershed is determined to be 3; and the associated bifurcation ratio, RB, is determined to be 4. a) Estimate the average number of streams of each order (1 through 3) for this basin. b) What is the total length of all streams for this basin? c) Define drainage density and average length of overland flow.
|
Stream Order, w |
Number of Streams, Nw |
Average Length (km) |
|
1 |
|
1.0 |
|
2 |
|
2.0 |
|
3 |
|
8.0 |
a) Use Horton's Law of Stream Numbers:
W = 3; then:
N1 = (4)(3-1) = 16.
N2 = (4)(3-2) = 4.
N3 = (4)(3-3) = 1.
b)
LT = 16 (1 km) + 4 (2 km) + 1 (8 km) = 32 km
c) Drainage Density is a measure of the average length of streams per unit area drained. It is a fundamental scale of the basin. Dd = LT/ABasin.
Average Length of Overland Flow, Lo is a measure of the average length that overland flow must travel before reaching a channel. Lo = 1/(2 Dd).
Problem 3 (25 points) The 1-h UH for a given watershed is tabulated below.
|
Time
(h) |
1 |
2 |
3 |
4 |
5 |
6 |
|
UH
(m3/s/cm) |
40 |
80 |
120 |
100 |
60 |
20 |
a) What is the area of the basin?
ABasin (0.01 m) = Vol. U.H.
Vol. U.H. = (40 + 80 + 120 + 100 + 60 + 20) m3/s 3600 s = 1'512,000 m3
Thus, ABasin
= 1'512,000 m3 / 0.01 m = 151'200,000 m2 = 151.2 km2
b) What is the DRH that would be observed from the following ERH?
|
Time
(h) |
0-1 |
1-2 |
2-3 |
|
ERH
(cm/h) |
0.25 |
1.0 |
0.75 |
P1 = 0.25 cm; P2 = 1.0 cm; P3 = 0.75 cm
|
UH (m3/s/cm) |
P1*UH (m3/s) |
P2*UH (m3/s) |
P3*UH (m3/s) |
DRH (m3/s) |
|
0 |
0 |
|
|
0 |
|
40 |
10 |
0 |
|
10 |
|
80 |
20 |
40 |
0 |
60 |
|
120 |
30 |
80 |
30 |
140 |
|
100 |
25 |
120 |
60 |
205 |
|
60 |
15 |
100 |
90 |
205 |
|
20 |
5 |
60 |
75 |
140 |
|
|
0 |
20 |
45 |
65 |
|
|
0 |
0 |
15 |
15 |
|
|
0 |
0 |
0 |
0 |
c) Define Unit Hydrograph and S-Hydrograph.
A Unit Hydrograph is the response of a basin to a unit volume of effective precipitation uniformly distributed over the basin and of effective duration D. The UH is labeled by the duration, D, of the effective precipitation.
An S-Hydrograph is the response of a basin to an infinite duration effective precipitation event of constant intensity 1/D and corresponding to the cumulative hydrograph volume from a D-UH. It is used in obtaining a D'-UH from a D-UH.